Given:
The 17th term of an A.P. is 5 more than twice its 8th term and the 11th term of the A.P. is 43.
To find:
the nth term.
Solution:
Given, a17 = 5 + 2(a8) ......(i)
And a11 = 43
As an = a + (n -1) d
For n = 11,
a + 10d = 43a = 43 - 10d ......(ii)
For n = 8, a8 = a + 7d
For n =17,
a17 = a + 16d
Put the value of a from (ii)
⇒ a17= 43 – 10d + 16d
⇒ a17 = 43 + 6d
Putting the value of a8 and a17 in (i), we get
⇒ 43 + 6d = 5 + 2(a + 7d)
⇒ 43 + 6d = 5 + 2a + 14d
⇒ 43 – 5 = 2a + 14d – 6d
⇒ 38 = 2a + 8d
⇒ 38 = 2(43 – 10d) + 8d [from (ii)]
⇒ 38 = 86 – 20d + 8d
⇒ 38 = 86 – 12d
⇒ 12d = 86 – 38
⇒ 12d = 48
⇒ d = 4
From (ii),
a = 43 – 10d
⇒ 43 – (10 × 4) = 43 – 40 = 3
We know, nth term of A.P., an = a + (n -1) d
an = 3 + (n – 1) 4
an = 3 + 4n – 4 an = 4n – 1
Hence, nth term is 4n – 1.